Maximize sum of topmost elements of S stacks by popping at most N elements
Given S stacks of length M, the task is to maximize the sum of elements at the top of each stack by popping at most N elements.
Example:
Input: S = 1, N = 3, stacks = { 5, 1, 2, 8, 9 }
Output: 8
Explanation:
Maximum 3 elements can be removed.
The current element at the top of the stack is 5.
On removal of 5, the new element at the top is 1.
On removal of 1, the new element at the top is 2.
On removal of 2, the new element at the top is 8.
No further pop operation is allowed.
Hence, the maximum possible value at the top of the stack is 8.
Input: S = 2, N = 2, stacks = { { 2, 6, 4, 5}, {1, 6, 15, 10} }
Output: 17
Explanation:
Current sum of the elements at the top = 2 + 1 = 3.
Popping 1 from top of the second stack only makes the sum 8 (5 + 2 = 8)
Popping 2 from the top of the second stack only makes the sum 7 (6 + 1).
Popping both 1 and 2 from the top of each stack makes the sum 12 (6 + 6).
Popping 2 and 6 from the first stack makes the sum 5 (4 + 1).
Popping 1 and 6 from the second stack leaves 15 as the element at the top.
Hence, the sum of elements at the top of the two stacks is maximized (15 + 2 = 17).
Approach: This problem can be reduced to a 0/1 Knapsack problem. To solve the problem, follow the steps below:
- Create a 2D table dp[][] with (S + 1) rows and (N + 1) columns. At every index dp[i][j], store the maximum sum possible by popping j elements up to the ith stack.
- Initialize all indices dp[][] by 0.
- Iterate over each stack from i = 0 to S – 1
- Now, for every ith stack, calculate the maximum possible sum by popping j (1 to N) elements..
- These j elements can be selected from all the i stacks already visited. Hence, dp[i+1][j] stores the maximum of stacks[i][k] + dp[i][j – k] for all values of k ranging from 0 to min(j, size of stack). The relation stacks[i][k] + dp[i][j-k] denotes the sum obtained by popping k elements from the current ith stack and maximum sum possible by popping j – k elements from the already visited stacks.
- Once, done for all i stacks, find the maximum of dp[S][i] for all i in range [1, N – 1].
- The maximum value obtained at the previous step is the required answer.
Below code is the implementation of the above approach:
C++
// C++ Program to maximize the // sum of top of the stack // values of S stacks by popping // at most N elements #include <bits/stdc++.h> using namespace std; // Function for computing the // maximum sum at the top of // the stacks after popping at // most N elements from S stack int maximumSum( int S, int M, int N, vector<vector< int > >& stacks) { // Constructing a dp matrix // of dimensions (S+1) x (N+1) int dp[S + 1][N + 1]; // Initialize all states memset (dp, INT_MIN, sizeof (dp)); // Loop over all i stacks for ( int i = 0; i < S; i++) { for ( int j = 0; j <= N; j++) { for ( int k = 0; k <= min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from current stack and // j - k elements from all // previous stacks combined dp[i + 1][j] = max(dp[i + 1][j], stacks[i][k] + dp[i][j - k]); } } } // Store the maximum sum of // popping N elements across // all stacks int result = INT_MIN; for ( int i = 0; i <= N; i++) { result = max(result, dp[S][i]); } // dp[S][N] has the maximum sum return result; } // Driver Program int main() { // Number of stacks int S = 2; // Length of each stack int M = 4; vector<vector< int > > stacks = { { 2, 6, 4, 5 }, { 1, 6, 15, 10 } }; // Maximum elements that // can be popped int N = 3; cout << maximumSum(S, M, N, stacks); return 0; } |
Java
// Java Program to maximize the // sum of top of the stack // values of S stacks by popping // at most N elements import java.util.*; class GFG{ // Function for computing the // maximum sum at the top of // the stacks after popping at // most N elements from S stack static int maximumSum( int S, int M, int N, int [][]stacks) { // Constructing a dp matrix // of dimensions (S+1) x (N+1) int [][]dp = new int [S + 1 ][N + 1 ]; // Loop over all i stacks for ( int i = 0 ; i < S; i++) { for ( int j = 0 ; j <= N; j++) { for ( int k = 0 ; k <= Math.min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from current stack and // j - k elements from all // previous stacks combined dp[i + 1 ][j] = Math.max(dp[i + 1 ][j], stacks[i][k] + dp[i][j - k]); } } } // Store the maximum sum of // popping N elements across // all stacks int result = Integer.MIN_VALUE; for ( int i = 0 ; i <= N; i++) { result = Math.max(result, dp[S][i]); } // dp[S][N] has the maximum sum return result; } // Driver Program public static void main(String[] args) { // Number of stacks int S = 2 ; // Length of each stack int M = 4 ; int [][]stacks = {{ 2 , 6 , 4 , 5 }, { 1 , 6 , 15 , 10 }}; // Maximum elements that // can be popped int N = 3 ; System.out.print(maximumSum(S, M, N, stacks)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to maximize the # sum of top of the stack values # of S stacks by popping at most # N element import sys # Function for computing the # maximum sum at the top of # the stacks after popping at # most N elements from S stack def maximumSum(S, M, N, stacks): # Constructing a dp matrix # of dimensions (S+1) x (N+1) dp = [[ 0 for x in range (N + 1 )] for y in range (S + 1 )] # Loop over all i stacks for i in range (S): for j in range (N + 1 ): for k in range ( min (j, M) + 1 ): # Store the maximum of # popping j elements # up to the current stack # by popping k elements # from current stack and # j - k elements from all # previous stacks combined dp[i + 1 ][j] = max (dp[i + 1 ][j], stacks[i][k] + dp[i][j - k]) # Store the maximum sum of # popping N elements across # all stacks result = - sys.maxsize - 1 for i in range (N + 1 ): result = max (result, dp[S][i]) # dp[S][N] has the maximum sum return result # Driver code if __name__ = = "__main__" : # Number of stacks S = 2 # Length of each stack M = 4 stacks = [ [ 2 , 6 , 4 , 5 ], [ 1 , 6 , 15 , 10 ] ] # Maximum elements that # can be popped N = 3 print (maximumSum(S, M, N, stacks)) # This code is contributed by chitranayal |
C#
// C# program to maximize the sum // of top of the stack values of // S stacks by popping at most N // elements using System; class GFG{ // Function for computing the // maximum sum at the top of // the stacks after popping at // most N elements from S stack static int maximumSum( int S, int M, int N, int [,]stacks) { // Constructing a dp matrix // of dimensions (S+1) x (N+1) int [,]dp = new int [S + 1, N + 1]; // Loop over all i stacks for ( int i = 0; i < S; i++) { for ( int j = 0; j <= N; j++) { for ( int k = 0; k <= Math.Min(j, M); k++) { // Store the maximum of popping // j elements up to the current // stack by popping k elements // from current stack and // j - k elements from all // previous stacks combined dp[i + 1, j] = Math.Max(dp[i + 1, j], stacks[i, k] + dp[i, j - k]); } } } // Store the maximum sum of // popping N elements across // all stacks int result = int .MinValue; for ( int i = 0; i <= N; i++) { result = Math.Max(result, dp[S, i]); } // dp[S,N] has the maximum sum return result; } // Driver code public static void Main(String[] args) { // Number of stacks int S = 2; // Length of each stack int M = 4; int [,]stacks = { { 2, 6, 4, 5 }, { 1, 6, 15, 10 } }; // Maximum elements that // can be popped int N = 3; Console.Write(maximumSum(S, M, N, stacks)); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript Program to maximize the // sum of top of the stack // values of S stacks by popping // at most N elements // Function for computing the // maximum sum at the top of // the stacks after popping at // most N elements from S stack function maximumSum(S, M, N, stacks) { // Constructing a dp matrix // of dimensions (S+1) x (N+1) var dp = Array.from(Array(S+1), ()=> Array(N+1).fill(0)); // Loop over all i stacks for ( var i = 0; i < S; i++) { for ( var j = 0; j <= N; j++) { for ( var k = 0; k <= Math.min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from current stack and // j - k elements from all // previous stacks combined dp[i + 1][j] = Math.max(dp[i + 1][j], stacks[i][k] + dp[i][j - k]); } } } // Store the maximum sum of // popping N elements across // all stacks var result = -1000000000; for ( var i = 0; i <= N; i++) { result = Math.max(result, dp[S][i]); } // dp[S][N] has the maximum sum return result; } // Driver Program // Number of stacks var S = 2; // Length of each stack var M = 4; var stacks = [ [ 2, 6, 4, 5 ], [ 1, 6, 15, 10 ] ]; // Maximum elements that // can be popped var N = 3; document.write( maximumSum(S, M, N, stacks)); </script> |
21
Time complexity: O( S*(M + N * (min(N, M))
Auxiliary Space: O(S*N), for dp array
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size N+1.
- Set a base case by initializing the values of DP .
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Now Create a temporary 1d vector new_dp used to store the current values from previous computations.
- After every iteration assign the value of new_dp to dp for further iteration.
- At last return and print the final answer stored in dp[N].
Implementation:
C++
#include <bits/stdc++.h> using namespace std; // Function for computing the // maximum sum at the top of // the stacks after popping at // most N elements from S stack int maximumSum( int S, int M, int N, vector<vector< int > >& stacks) { // Constructing a dp vector of size N+1 vector< int > dp(N + 1, INT_MIN); // Initialize the 0th index dp[0] = 0; // Loop over all i stacks for ( int i = 0; i < S; i++) { vector< int > new_dp(N + 1, INT_MIN); for ( int j = 0; j <= N; j++) { for ( int k = 0; k <= min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from current stack and // j - k elements from all // previous stacks combined new_dp[j] = max(new_dp[j], stacks[i][k] + dp[j - k]); } } dp = new_dp; } // dp[N] has the maximum sum return dp[N]; } // Driver Program int main() { // Number of stacks int S = 2; // Length of each stack int M = 4; vector<vector< int > > stacks = { { 2, 6, 4, 5 }, { 1, 6, 15, 10 } }; // Maximum elements that // can be popped int N = 3; cout << maximumSum(S, M, N, stacks); return 0; } // --- by bhardwajji |
Java
import java.util.*; public class Main { static int maximumSum( int S, int M, int N, List<List<Integer> > stacks) { // Constructing a dp vector of size N+1 List<Integer> dp = new ArrayList<>( Collections.nCopies(N + 1 , Integer.MIN_VALUE)); // Initialize the 0th index dp.set( 0 , 0 ); // Loop over all i stacks for ( int i = 0 ; i < S; i++) { List<Integer> new_dp = new ArrayList<>(Collections.nCopies( N + 1 , Integer.MIN_VALUE)); for ( int j = 0 ; j <= N; j++) { for ( int k = 0 ; k <= Math.min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from current stack and // j - k elements from all // previous stacks combined new_dp.set( j, Math.max(new_dp.get(j), stacks.get(i).get(k) + dp.get(j - k))); } } dp = new_dp; } // dp[N] has the maximum sum return dp.get(N); } public static void main(String[] args) { // Number of stacks int S = 2 ; // Length of each stack int M = 4 ; List<List<Integer> > stacks = new ArrayList<>(); stacks.add(Arrays.asList( 2 , 6 , 4 , 5 )); stacks.add(Arrays.asList( 1 , 6 , 15 , 10 )); // Maximum elements that can be popped int N = 3 ; System.out.println(maximumSum(S, M, N, stacks)); } } |
Python3
from typing import List # Function for computing the # maximum sum at the top of # the stacks after popping at # most N elements from S stack def maximumSum(S: int , M: int , N: int , stacks: List [ List [ int ]]) - > int : # Constructing a dp vector of size N+1 dp = [ float ( '-inf' )] * (N + 1 ) # Initialize the 0th index dp[ 0 ] = 0 # Loop over all i stacks for i in range (S): new_dp = [ float ( '-inf' )] * (N + 1 ) for j in range (N + 1 ): for k in range ( min (j, M) + 1 ): # Store the maximum of # popping j elements # up to the current stack # by popping k elements # from current stack and # j - k elements from all # previous stacks combined new_dp[j] = max (new_dp[j], stacks[i][k] + dp[j - k]) dp = new_dp # dp[N] has the maximum sum return dp[N] # Driver Program if __name__ = = '__main__' : # Number of stacks S = 2 # Length of each stack M = 4 stacks = [ [ 2 , 6 , 4 , 5 ], [ 1 , 6 , 15 , 10 ] ] # Maximum elements that # can be popped N = 3 print (maximumSum(S, M, N, stacks)) |
C#
using System; using System.Collections.Generic; public class GFG { static int MaximumSum( int S, int M, int N, List<List< int >> stacks) { // Constructing a dp vector of size N+1 List< int > dp = new List< int >( new int [N + 1]); // Initialize the 0th index dp[0] = 0; // Loop over all i stacks for ( int i = 0; i < S; i++) { List< int > new_dp = new List< int >( new int [N + 1]); for ( int j = 0; j <= N; j++) { for ( int k = 0; k <= Math.Min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from the current stack and // j - k elements from all // previous stacks combined new_dp[j] = Math.Max(new_dp[j], stacks[i][k] + dp[j - k]); } } dp = new_dp; } // dp[N] has the maximum sum return dp[N]; } public static void Main( string [] args) { // Number of stacks int S = 2; // Length of each stack int M = 4; List<List< int >> stacks = new List<List< int >>(); stacks.Add( new List< int > { 2, 6, 4, 5 }); stacks.Add( new List< int > { 1, 6, 15, 10 }); // Maximum elements that can be popped int N = 3; Console.WriteLine(MaximumSum(S, M, N, stacks)); } } //This code is contributed by aeroabrar_31 |
Javascript
// Function for computing the // maximum sum at the top of // the stacks after popping at // most N elements from S stack function maximumSum(S, M, N, stacks) { // Constructing a dp vector of size N+1 let dp = new Array(N + 1).fill(Number.MIN_SAFE_INTEGER); // Initialize the 0th index dp[0] = 0; // Loop over all i stacks for (let i = 0; i < S; i++) { let new_dp = new Array(N + 1).fill(Number.MIN_SAFE_INTEGER); for (let j = 0; j <= N; j++) { for (let k = 0; k <= Math.min(j, M); k++) { // Store the maximum of // popping j elements // up to the current stack // by popping k elements // from current stack and // j - k elements from all // previous stacks combined new_dp[j] = Math.max(new_dp[j], stacks[i][k] + dp[j - k]); } } dp = new_dp; } // dp[N] has the maximum sum return dp[N]; } // Driver Program // Number of stacks let S = 2; // Length of each stack let M = 4; let stacks = [ [2, 6, 4, 5], [1, 6, 15, 10] ]; // Maximum elements that // can be popped let N = 3; console.log(maximumSum(S, M, N, stacks)); |
Output:
21
Time complexity: O( S*(M + N * (min(N, M))
Auxiliary Space: O(N)